he Himalayas now reach an elevation of 8.8 km, and radiometric dating suggests that their uplift began about 45 million years ago. Assuming a constant rate of uplift, how fast did the Himalayas rise?

v = 6.20 10⁻¹² m / s,   v =  1.96 10⁻²  cm/year

Explanation:

This is a kinematics problem, specifically of uniform motion

        v = d / t

let's reduce the magnitudes to the SI system

      d = 8.8 km (1000 m / 1 km) = 8.8 10³ m

      t = 45 10⁶ years (365 day / 1 year) (24 h / 1 day) (3600 s / 1h)

       t = 1,419 10¹⁵ s

let's calculate the speed

        v = 8.8 10³ / 1.1419 10¹⁵

         v = 6.20 10⁻¹² m / s

This is the correct unit of the SI system, but it is more common to give the value in centimeters per year, so let's reduce the value

       v = 8.8 105 / 45 106

       v =  1.96 10⁻²  cm/year

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever